{"id":395,"date":"2021-02-04T01:58:16","date_gmt":"2021-02-04T01:58:16","guid":{"rendered":"https:\/\/courses.lumenlearning.com\/calculus1\/?post_type=chapter&#038;p=395"},"modified":"2022-03-16T05:43:16","modified_gmt":"2022-03-16T05:43:16","slug":"linear-approximation-of-a-function-at-a-point","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/calculus1\/chapter\/linear-approximation-of-a-function-at-a-point\/","title":{"raw":"Linear Approximation of a Function at a Point","rendered":"Linear Approximation of a Function at a Point"},"content":{"raw":"<div class=\"textbox learning-objectives\">\r\n<h3>Learning Outcomes<\/h3>\r\n<ul>\r\n \t<li>Describe the linear approximation to a function at a point.<\/li>\r\n \t<li>Write the linearization of a given function.<\/li>\r\n<\/ul>\r\n<\/div>\r\n<div id=\"fs-id1165042639971\" class=\"bc-section section\">\r\n<p id=\"fs-id1165043106586\">Consider a function [latex]f[\/latex] that is differentiable at a point [latex]x=a[\/latex]. Recall that the tangent line to the graph of [latex]f[\/latex] at [latex]a[\/latex] is given by the equation<\/p>\r\n\r\n<div id=\"fs-id1165042965164\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]y=f(a)+f^{\\prime}(a)(x-a)[\/latex]<\/div>\r\n<div><\/div>\r\nThis is simply derived from the point-slope form of the equation of a line\u00a0[latex]y-{y}_{1}=m\\left(x-{x}_{1}\\right)[\/latex] by adding\u00a0 [latex]{y}_{1}[\/latex] to both sides!\r\n\r\nFor example, consider the function [latex]f(x)=\\frac{1}{x}[\/latex] at [latex]a=2[\/latex]. Since [latex]f[\/latex] is differentiable at [latex]x=2[\/latex] and [latex]f^{\\prime}(x)=-\\frac{1}{x^2}[\/latex], we see that [latex]f^{\\prime}(2)=-\\frac{1}{4}[\/latex]. Therefore, the tangent line to the graph of [latex]f[\/latex] at [latex]a=2[\/latex] is given by the equation\r\n<div id=\"fs-id1165043259941\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]y=\\dfrac{1}{2}-\\dfrac{1}{4}(x-2)[\/latex]<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1165042514596\">Figure 1a shows a graph of [latex]f(x)=\\frac{1}{x}[\/latex] along with the tangent line to [latex]f[\/latex] at [latex]x=2[\/latex]. Note that for [latex]x[\/latex] near 2, the graph of the tangent line is close to the graph of [latex]f[\/latex]. As a result, we can use the equation of the tangent line to approximate [latex]f(x)[\/latex] for [latex]x[\/latex] near 2. For example, if [latex]x=2.1[\/latex], the [latex]y[\/latex] value of the corresponding point on the tangent line is<\/p>\r\n\r\n<div id=\"fs-id1165043429657\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]y=\\dfrac{1}{2}-\\dfrac{1}{4}(2.1-2)=0.475[\/latex]<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1165042613079\">The actual value of [latex]f(2.1)[\/latex] is given by<\/p>\r\n\r\n<div id=\"fs-id1165043306568\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]f(2.1)=\\dfrac{1}{2.1}\\approx 0.47619[\/latex]<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1165043157784\">Therefore, the tangent line gives us a fairly good approximation of [latex]f(2.1)[\/latex] (Figure 1b). However, note that for values of [latex]x[\/latex] far from 2, the equation of the tangent line does not give us a good approximation. For example, if [latex]x=10[\/latex], the [latex]y[\/latex]-value of the corresponding point on the tangent line is<\/p>\r\n\r\n<div id=\"fs-id1165042634904\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]y=\\frac{1}{2}-\\dfrac{1}{4}(10-2)=\\dfrac{1}{2}-2=-1.5[\/latex],<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1165043187581\">whereas the value of the function at [latex]x=10[\/latex] is [latex]f(10)=0.1[\/latex].<\/p>\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"851\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/2332\/2018\/01\/11210745\/CNX_Calc_Figure_04_02_006.jpg\" alt=\"This figure has two parts a and b. In figure a, the line f(x) = 1\/x is shown with its tangent line at x = 2. In figure b, the area near the tangent point is blown up to show how good of an approximation the tangent is near x = 2.\" width=\"851\" height=\"462\" \/> Figure 1. (a) The tangent line to [latex]f(x)=\\frac{1}{x}[\/latex] at [latex]x=2[\/latex] provides a good approximation to [latex]f[\/latex] for [latex]x[\/latex] near 2. (b) At [latex]x=2.1[\/latex], the value of [latex]y[\/latex] on the tangent line to [latex]f(x)=\\frac{1}{x}[\/latex] is 0.475. The actual value of [latex]f(2.1)[\/latex] is [latex]\\frac{1}{2.1}[\/latex], which is approximately 0.47619.[\/caption]\r\n<p id=\"fs-id1165043067518\">In general, for a differentiable function [latex]f[\/latex], the equation of the tangent line to [latex]f[\/latex] at [latex]x=a[\/latex] can be used to approximate [latex]f(x)[\/latex] for [latex]x[\/latex] near [latex]a[\/latex]. Therefore, we can write<\/p>\r\n\r\n<div id=\"fs-id1165042333160\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]f(x)\\approx f(a)+f^{\\prime}(a)(x-a)[\/latex] for [latex]x[\/latex] near [latex]a[\/latex]<\/div>\r\nWe call the linear function\r\n<div id=\"fs-id1165043306789\" class=\"equation\" style=\"text-align: center;\">[latex]L(x)=f(a)+f^{\\prime}(a)(x-a)[\/latex]<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1165043001981\">the <strong>linear approximation<\/strong>, or <strong>tangent line approximation<\/strong>, of [latex]f[\/latex] at [latex]x=a[\/latex]. This function [latex]L[\/latex] is also known as the linearization of [latex]f[\/latex] at [latex]x=a[\/latex].<\/p>\r\n<p id=\"fs-id1165042955335\">To show how useful the linear approximation can be, we look at how to find the linear approximation for [latex]f(x)=\\sqrt{x}[\/latex] at [latex]x=9[\/latex].<\/p>\r\n\r\n<div id=\"fs-id1165043051419\" class=\"textbook exercises\">\r\n<h3>Example: Linear Approximation of [latex]\\sqrt{x}[\/latex]<\/h3>\r\n<p id=\"fs-id1165043257102\">Find the linear approximation of [latex]f(x)=\\sqrt{x}[\/latex] at [latex]x=9[\/latex] and use the approximation to estimate [latex]\\sqrt{9.1}[\/latex].<\/p>\r\n[reveal-answer q=\"fs-id1165043351836\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1165043351836\"]\r\n<p id=\"fs-id1165043351836\">Since we are looking for the linear approximation at [latex]x=9[\/latex], using the tangent line approximation, we know the linear approximation is given by<\/p>\r\n\r\n<div id=\"fs-id1165042880023\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]L(x)=f(9)+f^{\\prime}(9)(x-9)[\/latex].<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1165043429300\">We need to find [latex]f(9)[\/latex] and [latex]f^{\\prime}(9)[\/latex].<\/p>\r\n\r\n<div id=\"fs-id1165043001328\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\begin{array}{lll} f(x)=\\sqrt{x}&amp; \\Rightarrow &amp; f(9)=\\sqrt{9}=3 \\\\ f^{\\prime}(x)=\\frac{1}{2\\sqrt{x}}&amp; \\Rightarrow &amp; f^{\\prime}(9)=\\frac{1}{2\\sqrt{9}}=\\frac{1}{6} \\end{array}[\/latex]<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1165043106851\">Therefore, the linear approximation is given by Figure 2.<\/p>\r\n\r\n<div id=\"fs-id1165042370716\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]L(x)=3+\\frac{1}{6}(x-9)[\/latex]<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1165042332089\">Using the linear approximation, we can estimate [latex]\\sqrt{9.1}[\/latex] by writing<\/p>\r\n\r\n<div id=\"fs-id1165042922906\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\sqrt{9.1}=f(9.1)\\approx L(9.1)=3+\\frac{1}{6}(9.1-9)\\approx 3.0167[\/latex].<\/div>\r\n<div><\/div>\r\n[caption id=\"\" align=\"aligncenter\" width=\"487\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/2332\/2018\/01\/11210748\/CNX_Calc_Figure_04_02_002.jpg\" alt=\"The function f(x) = the square root of x is shown with its tangent at (9, 3). The tangent appears to be a very good approximation from x = 6 to x = 12.\" width=\"487\" height=\"198\" \/> Figure 2. The local linear approximation to [latex]f(x)=\\sqrt{x}[\/latex] at [latex]x=9[\/latex] provides an approximation to [latex]f[\/latex] for [latex]x[\/latex] near 9.[\/caption]\r\n<h4>Analysis<\/h4>\r\n<p id=\"fs-id1165042711813\">Using a calculator, the value of [latex]\\sqrt{9.1}[\/latex] to four decimal places is 3.0166. The value given by the linear approximation, 3.0167, is very close to the value obtained with a calculator, so it appears that using this linear approximation is a good way to estimate [latex]\\sqrt{x}[\/latex], at least for [latex]x[\/latex] near 9. At the same time, it may seem odd to use a linear approximation when we can just push a few buttons on a calculator to evaluate [latex]\\sqrt{9.1}[\/latex]. However, how does the calculator evaluate [latex]\\sqrt{9.1}[\/latex]? The calculator uses an approximation! In fact, calculators and computers use approximations all the time to evaluate mathematical expressions; they just use higher-degree approximations.<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\nWatch the following video to see the worked solution to Example: Linear Approximation of [latex]\\sqrt{x}[\/latex].\r\n\r\n<center><iframe title=\"YouTube video player\" src=\"https:\/\/www.youtube.com\/embed\/R5C9K3oedGQ?controls=0&amp;start=128&amp;end=304&amp;autoplay=0\" width=\"750\" height=\"450\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/center>[reveal-answer q=\"266834\"]Closed Captioning and Transcript Information for Video[\/reveal-answer]\r\n[hidden-answer a=\"266834\"]For closed captioning, open the video on its original page by clicking the Youtube logo in the lower right-hand corner of the video display. In YouTube, the video will begin at the same starting point as this clip, but will continue playing until the very end.\r\n\r\nYou can view the <a href=\"https:\/\/oerfiles.s3.us-west-2.amazonaws.com\/Calculus\/Calculus1+Videos\/4.2LinearApproximationsAndDifferentials128to304_transcript.txt\" target=\"_blank\" rel=\"noopener\">transcript for this segmented clip of \"4.2 Linear Approximations and Differentials\" here (opens in new window)<\/a>.[\/hidden-answer]\r\n<div id=\"fs-id1165043105329\" class=\"textbook key-takeaways\">\r\n<h3>Try It<\/h3>\r\n<p id=\"fs-id1165042444934\">Find the local linear approximation to [latex]f(x)=\\sqrt[3]{x}[\/latex] at [latex]x=8[\/latex]. Use it to approximate [latex]\\sqrt[3]{8.1}[\/latex] to five decimal places.<\/p>\r\n[reveal-answer q=\"29087744\"]Hint[\/reveal-answer]\r\n[hidden-answer a=\"29087744\"]\r\n<p id=\"fs-id1165043321518\">[latex]L(x)=f(a)+f^{\\prime}(a)(x-a)[\/latex]<\/p>\r\n[\/hidden-answer]\r\n\r\n[reveal-answer q=\"fs-id1165042980470\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1165042980470\"]\r\n<p id=\"fs-id1165042980470\">[latex]L(x)=2+\\frac{1}{12}(x-8)[\/latex]; 2.00833<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div id=\"fs-id1165043100241\" class=\"textbook exercises\">\r\n<h3>Example: Linear Approximation of [latex] \\sin x[\/latex]<\/h3>\r\nFind the linear approximation of [latex]f(x)= \\sin x[\/latex] at [latex]x=\\dfrac{\\pi}{3}[\/latex] and use it to approximate [latex]\\sin (62^{\\circ})[\/latex].\r\n\r\n[reveal-answer q=\"fs-id1165043111804\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1165043111804\"]\r\n<p id=\"fs-id1165043111804\">First we note that since [latex]\\frac{\\pi}{3}[\/latex] rad is equivalent to [latex]60^{\\circ}[\/latex], using the linear approximation at [latex]x=\\pi \/3[\/latex] seems reasonable. The linear approximation is given by<\/p>\r\n\r\n<div id=\"fs-id1165043090235\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]L(x)=f(\\frac{\\pi}{3})+f^{\\prime}(\\frac{\\pi}{3})(x-\\frac{\\pi}{3})[\/latex].<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1165042551936\">We see that<\/p>\r\n\r\n<div id=\"fs-id1165043285221\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\begin{array}{lll}f(x)= \\sin x &amp; \\Rightarrow &amp; f(\\frac{\\pi}{3})= \\sin (\\frac{\\pi}{3})=\\frac{\\sqrt{3}}{2} \\\\ f^{\\prime}(x)= \\cos x &amp; \\Rightarrow &amp; f^{\\prime}(\\frac{\\pi}{3})= \\cos (\\frac{\\pi}{3})=\\frac{1}{2} \\end{array}[\/latex]<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1165043073754\">Therefore, the linear approximation of [latex]f[\/latex] at [latex]x=\\pi \/3[\/latex] is given by Figure 3.<\/p>\r\n\r\n<div id=\"fs-id1165042332414\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]L(x)=\\frac{\\sqrt{3}}{2}+\\frac{1}{2}(x-\\frac{\\pi}{3})[\/latex]<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1165043078487\">To estimate [latex] \\sin (62^{\\circ})[\/latex] using [latex]L[\/latex], we must first convert [latex]62^{\\circ}[\/latex] to radians. We have [latex]62^{\\circ}=\\frac{62\\pi}{180}[\/latex] radians, so the estimate for [latex] \\sin (62^{\\circ})[\/latex] is given by<\/p>\r\n\r\n<div id=\"fs-id1165043013778\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex] \\sin (62^{\\circ})=f(\\frac{62\\pi}{180})\\approx L(\\frac{62\\pi }{180})=\\frac{\\sqrt{3}}{2}+\\frac{1}{2}(\\frac{62\\pi }{180}-\\frac{\\pi }{3})=\\frac{\\sqrt{3}}{2}+\\frac{1}{2}(\\frac{2\\pi }{180})=\\frac{\\sqrt{3}}{2}+\\frac{\\pi }{180}\\approx 0.88348[\/latex].<\/div>\r\n<div><\/div>\r\n<div>[caption id=\"\" align=\"aligncenter\" width=\"731\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/2332\/2018\/01\/11210750\/CNX_Calc_Figure_04_02_003.jpg\" alt=\"The function f(x) = sin x is shown with its tangent at (\u03c0\/3, square root of 3 \/ 2). The tangent appears to be a very good approximation for x near \u03c0 \/ 3.\" width=\"731\" height=\"275\" \/> Figure 3. The linear approximation to [latex]f(x)= \\sin x[\/latex] at [latex]x=\\frac{\\pi}{3}[\/latex] provides an approximation to [latex] \\sin x[\/latex] for [latex]x[\/latex] near [latex]\\frac{\\pi}{3}.[\/latex][\/caption]<\/div>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\nWatch the following video to see the worked solution to Example: Linear Approximation of [latex] \\sin x[\/latex].\r\n\r\n<center><iframe title=\"YouTube video player\" src=\"https:\/\/www.youtube.com\/embed\/R5C9K3oedGQ?controls=0&amp;start=305&amp;end=437&amp;autoplay=0\" width=\"750\" height=\"450\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/center>[reveal-answer q=\"266833\"]Closed Captioning and Transcript Information for Video[\/reveal-answer]\r\n[hidden-answer a=\"266833\"]For closed captioning, open the video on its original page by clicking the Youtube logo in the lower right-hand corner of the video display. In YouTube, the video will begin at the same starting point as this clip, but will continue playing until the very end.\r\n\r\nYou can view the <a href=\"https:\/\/oerfiles.s3.us-west-2.amazonaws.com\/Calculus\/Calculus1+Videos\/4.2LinearApproximationsAndDifferentials305to437_transcript.txt\" target=\"_blank\" rel=\"noopener\">transcript for this segmented clip of \"4.2 Linear Approximations and Differentials\" here (opens in new window)<\/a>.[\/hidden-answer]\r\n<div id=\"fs-id1165042322345\" class=\"textbook key-takeaways\">\r\n<h3>Try It<\/h3>\r\n<p id=\"fs-id1165043060547\">Find the linear approximation for [latex]f(x)= \\cos x[\/latex] at [latex]x=\\dfrac{\\pi }{2}[\/latex].<\/p>\r\n[reveal-answer q=\"8834672\"]Hint[\/reveal-answer]\r\n[hidden-answer a=\"8834672\"]\r\n\r\n[latex]L(x)=f(a)+f^{\\prime}(a)(x-a)[\/latex]\r\n\r\n[\/hidden-answer]\r\n\r\n[reveal-answer q=\"902505\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"902505\"]\r\n\r\n[latex]L(x)=\u2212x+\\frac{\\pi}{2}[\/latex]\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<p id=\"fs-id1165042316225\">Linear approximations may be used in estimating roots and powers. In the next example, we find the linear approximation for [latex]f(x)=(1+x)^n[\/latex] at [latex]x=0[\/latex], which can be used to estimate roots and powers for real numbers near 1. The same idea can be extended to a function of the form [latex]f(x)=(m+x)^n[\/latex] to estimate roots and powers near a different number [latex]m[\/latex].<\/p>\r\n\r\n<div id=\"fs-id1165043161464\" class=\"textbook exercises\">\r\n<h3>Example: Approximating Roots and Powers<\/h3>\r\n<p id=\"fs-id1165042604945\">Find the linear approximation of [latex]f(x)=(1+x)^n[\/latex] at [latex]x=0[\/latex]. Use this approximation to estimate [latex](1.01)^3[\/latex].<\/p>\r\n[reveal-answer q=\"fs-id1165043396408\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1165043396408\"]\r\n<p id=\"fs-id1165043396408\">The linear approximation at [latex]x=0[\/latex] is given by<\/p>\r\n\r\n<div id=\"fs-id1165042710456\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]L(x)=f(0)+f^{\\prime}(0)(x-0)[\/latex].<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1165042553978\">Because<\/p>\r\n\r\n<div id=\"fs-id1165042369140\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\begin{array}{lll} f(x)=(1+x)^n &amp; \\Rightarrow &amp; f(0)=1 \\\\ f^{\\prime}(x)=n(1+x)^{n-1} &amp; \\Rightarrow &amp; f^{\\prime}(0)=n, \\end{array}[\/latex]<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1165043062849\">the linear approximation is given by Figure 4a.<\/p>\r\n\r\n<div id=\"fs-id1165042925702\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]L(x)=1+n(x-0)=1+nx[\/latex]<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1165043374164\">We can approximate [latex](1.01)^3[\/latex] by evaluating [latex]L(0.01)[\/latex] when [latex]n=3[\/latex]. We conclude that<\/p>\r\n\r\n<div id=\"fs-id1165042966729\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex](1.01)^3=f(1.01)\\approx L(1.01)=1+3(0.01)=1.03[\/latex].<\/div>\r\n<div><\/div>\r\n<div>[caption id=\"\" align=\"aligncenter\" width=\"814\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/2332\/2018\/01\/11210754\/CNX_Calc_Figure_04_02_007.jpg\" alt=\"This figure has two parts a and b. In figure a, the line f(x) = (1 + x)3 is shown with its tangent line at (0, 1). In figure b, the area near the tangent point is blown up to show how good of an approximation the tangent is near (0, 1).\" width=\"814\" height=\"387\" \/> Figure 4. (a) The linear approximation of [latex]f(x)[\/latex] at [latex]x=0[\/latex] is [latex]L(x)[\/latex]. (b) The actual value of [latex]1.01^3[\/latex] is 1.030301. The linear approximation of [latex]f(x)[\/latex] at [latex]x=0[\/latex] estimates [latex]1.01^3[\/latex] to be 1.03.[\/caption]<\/div>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div id=\"fs-id1165043286811\" class=\"textbook key-takeaways\">\r\n<h3>Try It<\/h3>\r\n<p id=\"fs-id1165043394826\">Find the linear approximation of [latex]f(x)=(1+x)^4[\/latex] at [latex]x=0[\/latex] without using the result from the preceding example.<\/p>\r\n[reveal-answer q=\"477002\"]Hint[\/reveal-answer]\r\n[hidden-answer a=\"477002\"]\r\n<p id=\"fs-id1165042964925\">[latex]f^{\\prime}(x)=4(1+x)^3[\/latex]<\/p>\r\n[\/hidden-answer]\r\n\r\n[reveal-answer q=\"fs-id1165043253559\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1165043253559\"]\r\n<p id=\"fs-id1165043253559\">[latex]L(x)=1+4x[\/latex]<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Try It<\/h3>\r\n[ohm_question]5182[\/ohm_question]\r\n\r\n<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Try It<\/h3>\r\n[ohm_question]5186[\/ohm_question]\r\n\r\n<\/div>","rendered":"<div class=\"textbox learning-objectives\">\n<h3>Learning Outcomes<\/h3>\n<ul>\n<li>Describe the linear approximation to a function at a point.<\/li>\n<li>Write the linearization of a given function.<\/li>\n<\/ul>\n<\/div>\n<div id=\"fs-id1165042639971\" class=\"bc-section section\">\n<p id=\"fs-id1165043106586\">Consider a function [latex]f[\/latex] that is differentiable at a point [latex]x=a[\/latex]. Recall that the tangent line to the graph of [latex]f[\/latex] at [latex]a[\/latex] is given by the equation<\/p>\n<div id=\"fs-id1165042965164\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]y=f(a)+f^{\\prime}(a)(x-a)[\/latex]<\/div>\n<div><\/div>\n<p>This is simply derived from the point-slope form of the equation of a line\u00a0[latex]y-{y}_{1}=m\\left(x-{x}_{1}\\right)[\/latex] by adding\u00a0 [latex]{y}_{1}[\/latex] to both sides!<\/p>\n<p>For example, consider the function [latex]f(x)=\\frac{1}{x}[\/latex] at [latex]a=2[\/latex]. Since [latex]f[\/latex] is differentiable at [latex]x=2[\/latex] and [latex]f^{\\prime}(x)=-\\frac{1}{x^2}[\/latex], we see that [latex]f^{\\prime}(2)=-\\frac{1}{4}[\/latex]. Therefore, the tangent line to the graph of [latex]f[\/latex] at [latex]a=2[\/latex] is given by the equation<\/p>\n<div id=\"fs-id1165043259941\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]y=\\dfrac{1}{2}-\\dfrac{1}{4}(x-2)[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1165042514596\">Figure 1a shows a graph of [latex]f(x)=\\frac{1}{x}[\/latex] along with the tangent line to [latex]f[\/latex] at [latex]x=2[\/latex]. Note that for [latex]x[\/latex] near 2, the graph of the tangent line is close to the graph of [latex]f[\/latex]. As a result, we can use the equation of the tangent line to approximate [latex]f(x)[\/latex] for [latex]x[\/latex] near 2. For example, if [latex]x=2.1[\/latex], the [latex]y[\/latex] value of the corresponding point on the tangent line is<\/p>\n<div id=\"fs-id1165043429657\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]y=\\dfrac{1}{2}-\\dfrac{1}{4}(2.1-2)=0.475[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1165042613079\">The actual value of [latex]f(2.1)[\/latex] is given by<\/p>\n<div id=\"fs-id1165043306568\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]f(2.1)=\\dfrac{1}{2.1}\\approx 0.47619[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1165043157784\">Therefore, the tangent line gives us a fairly good approximation of [latex]f(2.1)[\/latex] (Figure 1b). However, note that for values of [latex]x[\/latex] far from 2, the equation of the tangent line does not give us a good approximation. For example, if [latex]x=10[\/latex], the [latex]y[\/latex]-value of the corresponding point on the tangent line is<\/p>\n<div id=\"fs-id1165042634904\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]y=\\frac{1}{2}-\\dfrac{1}{4}(10-2)=\\dfrac{1}{2}-2=-1.5[\/latex],<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1165043187581\">whereas the value of the function at [latex]x=10[\/latex] is [latex]f(10)=0.1[\/latex].<\/p>\n<div style=\"width: 861px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/2332\/2018\/01\/11210745\/CNX_Calc_Figure_04_02_006.jpg\" alt=\"This figure has two parts a and b. In figure a, the line f(x) = 1\/x is shown with its tangent line at x = 2. In figure b, the area near the tangent point is blown up to show how good of an approximation the tangent is near x = 2.\" width=\"851\" height=\"462\" \/><\/p>\n<p class=\"wp-caption-text\">Figure 1. (a) The tangent line to [latex]f(x)=\\frac{1}{x}[\/latex] at [latex]x=2[\/latex] provides a good approximation to [latex]f[\/latex] for [latex]x[\/latex] near 2. (b) At [latex]x=2.1[\/latex], the value of [latex]y[\/latex] on the tangent line to [latex]f(x)=\\frac{1}{x}[\/latex] is 0.475. The actual value of [latex]f(2.1)[\/latex] is [latex]\\frac{1}{2.1}[\/latex], which is approximately 0.47619.<\/p>\n<\/div>\n<p id=\"fs-id1165043067518\">In general, for a differentiable function [latex]f[\/latex], the equation of the tangent line to [latex]f[\/latex] at [latex]x=a[\/latex] can be used to approximate [latex]f(x)[\/latex] for [latex]x[\/latex] near [latex]a[\/latex]. Therefore, we can write<\/p>\n<div id=\"fs-id1165042333160\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]f(x)\\approx f(a)+f^{\\prime}(a)(x-a)[\/latex] for [latex]x[\/latex] near [latex]a[\/latex]<\/div>\n<p>We call the linear function<\/p>\n<div id=\"fs-id1165043306789\" class=\"equation\" style=\"text-align: center;\">[latex]L(x)=f(a)+f^{\\prime}(a)(x-a)[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1165043001981\">the <strong>linear approximation<\/strong>, or <strong>tangent line approximation<\/strong>, of [latex]f[\/latex] at [latex]x=a[\/latex]. This function [latex]L[\/latex] is also known as the linearization of [latex]f[\/latex] at [latex]x=a[\/latex].<\/p>\n<p id=\"fs-id1165042955335\">To show how useful the linear approximation can be, we look at how to find the linear approximation for [latex]f(x)=\\sqrt{x}[\/latex] at [latex]x=9[\/latex].<\/p>\n<div id=\"fs-id1165043051419\" class=\"textbook exercises\">\n<h3>Example: Linear Approximation of [latex]\\sqrt{x}[\/latex]<\/h3>\n<p id=\"fs-id1165043257102\">Find the linear approximation of [latex]f(x)=\\sqrt{x}[\/latex] at [latex]x=9[\/latex] and use the approximation to estimate [latex]\\sqrt{9.1}[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1165043351836\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1165043351836\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1165043351836\">Since we are looking for the linear approximation at [latex]x=9[\/latex], using the tangent line approximation, we know the linear approximation is given by<\/p>\n<div id=\"fs-id1165042880023\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]L(x)=f(9)+f^{\\prime}(9)(x-9)[\/latex].<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1165043429300\">We need to find [latex]f(9)[\/latex] and [latex]f^{\\prime}(9)[\/latex].<\/p>\n<div id=\"fs-id1165043001328\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\begin{array}{lll} f(x)=\\sqrt{x}& \\Rightarrow & f(9)=\\sqrt{9}=3 \\\\ f^{\\prime}(x)=\\frac{1}{2\\sqrt{x}}& \\Rightarrow & f^{\\prime}(9)=\\frac{1}{2\\sqrt{9}}=\\frac{1}{6} \\end{array}[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1165043106851\">Therefore, the linear approximation is given by Figure 2.<\/p>\n<div id=\"fs-id1165042370716\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]L(x)=3+\\frac{1}{6}(x-9)[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1165042332089\">Using the linear approximation, we can estimate [latex]\\sqrt{9.1}[\/latex] by writing<\/p>\n<div id=\"fs-id1165042922906\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\sqrt{9.1}=f(9.1)\\approx L(9.1)=3+\\frac{1}{6}(9.1-9)\\approx 3.0167[\/latex].<\/div>\n<div><\/div>\n<div style=\"width: 497px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/2332\/2018\/01\/11210748\/CNX_Calc_Figure_04_02_002.jpg\" alt=\"The function f(x) = the square root of x is shown with its tangent at (9, 3). The tangent appears to be a very good approximation from x = 6 to x = 12.\" width=\"487\" height=\"198\" \/><\/p>\n<p class=\"wp-caption-text\">Figure 2. The local linear approximation to [latex]f(x)=\\sqrt{x}[\/latex] at [latex]x=9[\/latex] provides an approximation to [latex]f[\/latex] for [latex]x[\/latex] near 9.<\/p>\n<\/div>\n<h4>Analysis<\/h4>\n<p id=\"fs-id1165042711813\">Using a calculator, the value of [latex]\\sqrt{9.1}[\/latex] to four decimal places is 3.0166. The value given by the linear approximation, 3.0167, is very close to the value obtained with a calculator, so it appears that using this linear approximation is a good way to estimate [latex]\\sqrt{x}[\/latex], at least for [latex]x[\/latex] near 9. At the same time, it may seem odd to use a linear approximation when we can just push a few buttons on a calculator to evaluate [latex]\\sqrt{9.1}[\/latex]. However, how does the calculator evaluate [latex]\\sqrt{9.1}[\/latex]? The calculator uses an approximation! In fact, calculators and computers use approximations all the time to evaluate mathematical expressions; they just use higher-degree approximations.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<p>Watch the following video to see the worked solution to Example: Linear Approximation of [latex]\\sqrt{x}[\/latex].<\/p>\n<div style=\"text-align: center;\"><iframe loading=\"lazy\" title=\"YouTube video player\" src=\"https:\/\/www.youtube.com\/embed\/R5C9K3oedGQ?controls=0&amp;start=128&amp;end=304&amp;autoplay=0\" width=\"750\" height=\"450\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/div>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q266834\">Closed Captioning and Transcript Information for Video<\/span><\/p>\n<div id=\"q266834\" class=\"hidden-answer\" style=\"display: none\">For closed captioning, open the video on its original page by clicking the Youtube logo in the lower right-hand corner of the video display. In YouTube, the video will begin at the same starting point as this clip, but will continue playing until the very end.<\/p>\n<p>You can view the <a href=\"https:\/\/oerfiles.s3.us-west-2.amazonaws.com\/Calculus\/Calculus1+Videos\/4.2LinearApproximationsAndDifferentials128to304_transcript.txt\" target=\"_blank\" rel=\"noopener\">transcript for this segmented clip of &#8220;4.2 Linear Approximations and Differentials&#8221; here (opens in new window)<\/a>.<\/div>\n<\/div>\n<div id=\"fs-id1165043105329\" class=\"textbook key-takeaways\">\n<h3>Try It<\/h3>\n<p id=\"fs-id1165042444934\">Find the local linear approximation to [latex]f(x)=\\sqrt[3]{x}[\/latex] at [latex]x=8[\/latex]. Use it to approximate [latex]\\sqrt[3]{8.1}[\/latex] to five decimal places.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q29087744\">Hint<\/span><\/p>\n<div id=\"q29087744\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1165043321518\">[latex]L(x)=f(a)+f^{\\prime}(a)(x-a)[\/latex]<\/p>\n<\/div>\n<\/div>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1165042980470\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1165042980470\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1165042980470\">[latex]L(x)=2+\\frac{1}{12}(x-8)[\/latex]; 2.00833<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id1165043100241\" class=\"textbook exercises\">\n<h3>Example: Linear Approximation of [latex]\\sin x[\/latex]<\/h3>\n<p>Find the linear approximation of [latex]f(x)= \\sin x[\/latex] at [latex]x=\\dfrac{\\pi}{3}[\/latex] and use it to approximate [latex]\\sin (62^{\\circ})[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1165043111804\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1165043111804\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1165043111804\">First we note that since [latex]\\frac{\\pi}{3}[\/latex] rad is equivalent to [latex]60^{\\circ}[\/latex], using the linear approximation at [latex]x=\\pi \/3[\/latex] seems reasonable. The linear approximation is given by<\/p>\n<div id=\"fs-id1165043090235\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]L(x)=f(\\frac{\\pi}{3})+f^{\\prime}(\\frac{\\pi}{3})(x-\\frac{\\pi}{3})[\/latex].<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1165042551936\">We see that<\/p>\n<div id=\"fs-id1165043285221\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\begin{array}{lll}f(x)= \\sin x & \\Rightarrow & f(\\frac{\\pi}{3})= \\sin (\\frac{\\pi}{3})=\\frac{\\sqrt{3}}{2} \\\\ f^{\\prime}(x)= \\cos x & \\Rightarrow & f^{\\prime}(\\frac{\\pi}{3})= \\cos (\\frac{\\pi}{3})=\\frac{1}{2} \\end{array}[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1165043073754\">Therefore, the linear approximation of [latex]f[\/latex] at [latex]x=\\pi \/3[\/latex] is given by Figure 3.<\/p>\n<div id=\"fs-id1165042332414\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]L(x)=\\frac{\\sqrt{3}}{2}+\\frac{1}{2}(x-\\frac{\\pi}{3})[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1165043078487\">To estimate [latex]\\sin (62^{\\circ})[\/latex] using [latex]L[\/latex], we must first convert [latex]62^{\\circ}[\/latex] to radians. We have [latex]62^{\\circ}=\\frac{62\\pi}{180}[\/latex] radians, so the estimate for [latex]\\sin (62^{\\circ})[\/latex] is given by<\/p>\n<div id=\"fs-id1165043013778\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\sin (62^{\\circ})=f(\\frac{62\\pi}{180})\\approx L(\\frac{62\\pi }{180})=\\frac{\\sqrt{3}}{2}+\\frac{1}{2}(\\frac{62\\pi }{180}-\\frac{\\pi }{3})=\\frac{\\sqrt{3}}{2}+\\frac{1}{2}(\\frac{2\\pi }{180})=\\frac{\\sqrt{3}}{2}+\\frac{\\pi }{180}\\approx 0.88348[\/latex].<\/div>\n<div><\/div>\n<div>\n<div style=\"width: 741px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/2332\/2018\/01\/11210750\/CNX_Calc_Figure_04_02_003.jpg\" alt=\"The function f(x) = sin x is shown with its tangent at (\u03c0\/3, square root of 3 \/ 2). The tangent appears to be a very good approximation for x near \u03c0 \/ 3.\" width=\"731\" height=\"275\" \/><\/p>\n<p class=\"wp-caption-text\">Figure 3. The linear approximation to [latex]f(x)= \\sin x[\/latex] at [latex]x=\\frac{\\pi}{3}[\/latex] provides an approximation to [latex] \\sin x[\/latex] for [latex]x[\/latex] near [latex]\\frac{\\pi}{3}.[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<p>Watch the following video to see the worked solution to Example: Linear Approximation of [latex]\\sin x[\/latex].<\/p>\n<div style=\"text-align: center;\"><iframe loading=\"lazy\" title=\"YouTube video player\" src=\"https:\/\/www.youtube.com\/embed\/R5C9K3oedGQ?controls=0&amp;start=305&amp;end=437&amp;autoplay=0\" width=\"750\" height=\"450\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/div>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q266833\">Closed Captioning and Transcript Information for Video<\/span><\/p>\n<div id=\"q266833\" class=\"hidden-answer\" style=\"display: none\">For closed captioning, open the video on its original page by clicking the Youtube logo in the lower right-hand corner of the video display. In YouTube, the video will begin at the same starting point as this clip, but will continue playing until the very end.<\/p>\n<p>You can view the <a href=\"https:\/\/oerfiles.s3.us-west-2.amazonaws.com\/Calculus\/Calculus1+Videos\/4.2LinearApproximationsAndDifferentials305to437_transcript.txt\" target=\"_blank\" rel=\"noopener\">transcript for this segmented clip of &#8220;4.2 Linear Approximations and Differentials&#8221; here (opens in new window)<\/a>.<\/div>\n<\/div>\n<div id=\"fs-id1165042322345\" class=\"textbook key-takeaways\">\n<h3>Try It<\/h3>\n<p id=\"fs-id1165043060547\">Find the linear approximation for [latex]f(x)= \\cos x[\/latex] at [latex]x=\\dfrac{\\pi }{2}[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q8834672\">Hint<\/span><\/p>\n<div id=\"q8834672\" class=\"hidden-answer\" style=\"display: none\">\n<p>[latex]L(x)=f(a)+f^{\\prime}(a)(x-a)[\/latex]<\/p>\n<\/div>\n<\/div>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q902505\">Show Solution<\/span><\/p>\n<div id=\"q902505\" class=\"hidden-answer\" style=\"display: none\">\n<p>[latex]L(x)=\u2212x+\\frac{\\pi}{2}[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<p id=\"fs-id1165042316225\">Linear approximations may be used in estimating roots and powers. In the next example, we find the linear approximation for [latex]f(x)=(1+x)^n[\/latex] at [latex]x=0[\/latex], which can be used to estimate roots and powers for real numbers near 1. The same idea can be extended to a function of the form [latex]f(x)=(m+x)^n[\/latex] to estimate roots and powers near a different number [latex]m[\/latex].<\/p>\n<div id=\"fs-id1165043161464\" class=\"textbook exercises\">\n<h3>Example: Approximating Roots and Powers<\/h3>\n<p id=\"fs-id1165042604945\">Find the linear approximation of [latex]f(x)=(1+x)^n[\/latex] at [latex]x=0[\/latex]. Use this approximation to estimate [latex](1.01)^3[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1165043396408\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1165043396408\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1165043396408\">The linear approximation at [latex]x=0[\/latex] is given by<\/p>\n<div id=\"fs-id1165042710456\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]L(x)=f(0)+f^{\\prime}(0)(x-0)[\/latex].<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1165042553978\">Because<\/p>\n<div id=\"fs-id1165042369140\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\begin{array}{lll} f(x)=(1+x)^n & \\Rightarrow & f(0)=1 \\\\ f^{\\prime}(x)=n(1+x)^{n-1} & \\Rightarrow & f^{\\prime}(0)=n, \\end{array}[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1165043062849\">the linear approximation is given by Figure 4a.<\/p>\n<div id=\"fs-id1165042925702\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]L(x)=1+n(x-0)=1+nx[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1165043374164\">We can approximate [latex](1.01)^3[\/latex] by evaluating [latex]L(0.01)[\/latex] when [latex]n=3[\/latex]. We conclude that<\/p>\n<div id=\"fs-id1165042966729\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex](1.01)^3=f(1.01)\\approx L(1.01)=1+3(0.01)=1.03[\/latex].<\/div>\n<div><\/div>\n<div>\n<div style=\"width: 824px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/2332\/2018\/01\/11210754\/CNX_Calc_Figure_04_02_007.jpg\" alt=\"This figure has two parts a and b. In figure a, the line f(x) = (1 + x)3 is shown with its tangent line at (0, 1). In figure b, the area near the tangent point is blown up to show how good of an approximation the tangent is near (0, 1).\" width=\"814\" height=\"387\" \/><\/p>\n<p class=\"wp-caption-text\">Figure 4. (a) The linear approximation of [latex]f(x)[\/latex] at [latex]x=0[\/latex] is [latex]L(x)[\/latex]. (b) The actual value of [latex]1.01^3[\/latex] is 1.030301. The linear approximation of [latex]f(x)[\/latex] at [latex]x=0[\/latex] estimates [latex]1.01^3[\/latex] to be 1.03.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id1165043286811\" class=\"textbook key-takeaways\">\n<h3>Try It<\/h3>\n<p id=\"fs-id1165043394826\">Find the linear approximation of [latex]f(x)=(1+x)^4[\/latex] at [latex]x=0[\/latex] without using the result from the preceding example.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q477002\">Hint<\/span><\/p>\n<div id=\"q477002\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1165042964925\">[latex]f^{\\prime}(x)=4(1+x)^3[\/latex]<\/p>\n<\/div>\n<\/div>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1165043253559\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1165043253559\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1165043253559\">[latex]L(x)=1+4x[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p><iframe loading=\"lazy\" id=\"ohm5182\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=5182&theme=oea&iframe_resize_id=ohm5182&show_question_numbers\" width=\"100%\" height=\"150\"><\/iframe><\/p>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p><iframe loading=\"lazy\" id=\"ohm5186\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=5186&theme=oea&iframe_resize_id=ohm5186&show_question_numbers\" width=\"100%\" height=\"150\"><\/iframe><\/p>\n<\/div>\n\n\t\t\t <section class=\"citations-section\" role=\"contentinfo\">\n\t\t\t <h3>Candela Citations<\/h3>\n\t\t\t\t\t <div>\n\t\t\t\t\t\t <div id=\"citation-list-395\">\n\t\t\t\t\t\t\t <div class=\"licensing\"><div class=\"license-attribution-dropdown-subheading\">CC licensed content, Original<\/div><ul class=\"citation-list\"><li>4.2 Linear Approximations and Differentials. <strong>Authored by<\/strong>: Ryan Melton. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em><\/li><\/ul><div class=\"license-attribution-dropdown-subheading\">CC licensed content, Shared previously<\/div><ul class=\"citation-list\"><li>Calculus Volume 1. <strong>Authored by<\/strong>: Gilbert Strang, Edwin (Jed) Herman. <strong>Provided by<\/strong>: OpenStax. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"https:\/\/openstax.org\/details\/books\/calculus-volume-1\">https:\/\/openstax.org\/details\/books\/calculus-volume-1<\/a>. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by-nc-sa\/4.0\/\">CC BY-NC-SA: Attribution-NonCommercial-ShareAlike<\/a><\/em>. <strong>License Terms<\/strong>: Access for free at https:\/\/openstax.org\/books\/calculus-volume-1\/pages\/1-introduction<\/li><\/ul><\/div>\n\t\t\t\t\t\t <\/div>\n\t\t\t\t\t <\/div>\n\t\t\t <\/section>","protected":false},"author":17533,"menu_order":6,"template":"","meta":{"_candela_citation":"[{\"type\":\"cc\",\"description\":\"Calculus Volume 1\",\"author\":\"Gilbert Strang, Edwin (Jed) Herman\",\"organization\":\"OpenStax\",\"url\":\"https:\/\/openstax.org\/details\/books\/calculus-volume-1\",\"project\":\"\",\"license\":\"cc-by-nc-sa\",\"license_terms\":\"Access for free at https:\/\/openstax.org\/books\/calculus-volume-1\/pages\/1-introduction\"},{\"type\":\"original\",\"description\":\"4.2 Linear Approximations and Differentials\",\"author\":\"Ryan Melton\",\"organization\":\"\",\"url\":\"\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"\"}]","CANDELA_OUTCOMES_GUID":"","pb_show_title":"on","pb_short_title":"","pb_subtitle":"","pb_authors":[],"pb_section_license":""},"chapter-type":[],"contributor":[],"license":[],"class_list":["post-395","chapter","type-chapter","status-publish","hentry"],"part":48,"_links":{"self":[{"href":"https:\/\/courses.lumenlearning.com\/calculus1\/wp-json\/pressbooks\/v2\/chapters\/395","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/courses.lumenlearning.com\/calculus1\/wp-json\/pressbooks\/v2\/chapters"}],"about":[{"href":"https:\/\/courses.lumenlearning.com\/calculus1\/wp-json\/wp\/v2\/types\/chapter"}],"author":[{"embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/calculus1\/wp-json\/wp\/v2\/users\/17533"}],"version-history":[{"count":22,"href":"https:\/\/courses.lumenlearning.com\/calculus1\/wp-json\/pressbooks\/v2\/chapters\/395\/revisions"}],"predecessor-version":[{"id":4826,"href":"https:\/\/courses.lumenlearning.com\/calculus1\/wp-json\/pressbooks\/v2\/chapters\/395\/revisions\/4826"}],"part":[{"href":"https:\/\/courses.lumenlearning.com\/calculus1\/wp-json\/pressbooks\/v2\/parts\/48"}],"metadata":[{"href":"https:\/\/courses.lumenlearning.com\/calculus1\/wp-json\/pressbooks\/v2\/chapters\/395\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/courses.lumenlearning.com\/calculus1\/wp-json\/wp\/v2\/media?parent=395"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/calculus1\/wp-json\/pressbooks\/v2\/chapter-type?post=395"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/calculus1\/wp-json\/wp\/v2\/contributor?post=395"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/calculus1\/wp-json\/wp\/v2\/license?post=395"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}